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1:00 AM
@Sanjana auto means self-written. You can at most write a biography on him.
@SillyGoose there is never too much cake. Just eat a cake to make more space. Whatever Alice had in wonderland
 
 
3 hours later…
4:26 AM
H O N K
B A H
 
5:22 AM
honk
@ACuriousMind if i wanted to look more into this would it just be in a text on classical field theory?
I would like to see a side by side comparison of the classical: particle and field formalism; and of the quantum: particle and field formalism. and then also see what it means to treat the particle and field formalisms together at once. e.g. in an introductory E&M course we do deal with both fields and particles at once
it seems like there is a lot of the same structure being used, but for different purposes
 
Actually, I think you should not look for particle formalism in the classical sense. It is probably only going to make you feel far more confused. Instead, the standard treatment that makes sense would be more like an electrodynamics situation, whereby you solve for some EM field configuration, and then look at how some charged particle would be pushed along by them.
The reason why it would be a bad idea, is that, classically, there is no photon particle. There is only the continuous Maxwellian EM field.
If you want to really go even deeper, you could probably look up the fluid mechanics, particularly magnetohydrodynamics. That field of study is so difficult that you can literally read off and classify the dayum plasma instabilities from the equations, and still not get a Nobel prize.
You should be able to make do with EM fields, because those are 2-form fields that we separate into 1-forms, and that can be converted back to the vector-field formalism, that then allows you to do your Lie derivative and so forth. The differential geometry toolbox can be unleashed upon them.
 
5:43 AM
@naturallyInconsistent just a completely off tangent point, I think In MHD they use the first order magnetostatic approximation to decouple the Maxwell's equations
Which I always found weird because I thought plasmas were fast fast
 
I am not sure that they do that. However, I am aware that it is commonly the case that they would find that their computations are so difficult and computationally intensive that they would turn stuff off if they can. Solving matrix equations might well take too long and too much memory space. I just heard from a guy that they would simply implement Ampere-Maxwell Law by hand and thus eliminate the magnetic field variable. Just E and J will do. For some reason, it is unstable and less accurate
but the speedup is worth it
 
 
1 hour later…
6:53 AM
@naturallyInconsistent wait sorry what do you mean by "we are simply forced to consider this subtraction as the actual measurement result" ?
 
@naturallyInconsistent i love alice in wonderland
 
Well, we wanted $D_F(x-y)$ propagator to be the measurement result. It has almost all the behaviours correct, but obviously wrong in the "should be zero upon spacelike separation". The subtraction enforces the zero that we want, and so we make a swap and consider the commutator to be the measurement result.
 
swaggums
have you got the swaggums? @Relativisticcucumber
 
@SillyGoose eat gochujang
 
6:56 AM
swaggums?
 
@Relativisticcucumber 너를 먹을 거예요
 
@JohnRennie enters the room to ask about gochujang ass again
 
@Relativisticcucumber translating the absurdities found in maths into children's stories is fun for all
 
@naturallyInconsistent oh no i am confused about this calculation. because i thought tong is saying here that the problem is that the probability is not zero, which is the leakage in mention here?
but then i see the point you are referring to where it says the subtraction is zero...
 
@Relativisticcucumber Because I am not looking at the parts earlier than your picture quote post of Tong, I am going to have to assume some stuff about it. Assuming that right before this computation, Tong showed that a satisfactory measurement operator should commute upon spacelike separations, i.e. be confined within the light cone, and then Equation (2.91) showed that a propagator alone gives a leakage of probability amplitudes, so we are forced to consider the commutator as the measurement operator.
 
7:11 AM
@naturallyInconsistent oh no
 
? What is the problem?
 
so he introduces the propagator to say that its wrong?
 
In a sense, yes, but then it can be immediately fixed by taking the propagator and subtracting them
 
because i thought we actually do go on to use the propagator, but from what i am gathering from what you said, actually the commutator is what we should use?
so why not just use the commutator from the get go?
 
This is a part of QFT that is somewhat understandable whenever I study it, but then it is not so important that I remember it exactly, so here goes: I think later on you will learn to get the Feynman propagator directly, and then we would always use this trickery to get the correct answers. If you want the rigorous answer, you can just look up Wick's theorem, which is always what is being applied when we do these.
Like, the commutator is kinda awkward, and we would have to explain why we care about the commutator, leading back to this same silly computation. The result is always the same; the presentation here is probably easier to follow compared to the opposite.
I think this is but a small hiccup; the headaches of QFT are everywhere so I'm not about to move hell or high water on this topic. There are way worse bits to bang heads on walls about.
 
7:20 AM
@RyderRude $\frac{d}{dt} \hat{f}(t) = \frac{\partial \hat{f}}{\partial t} + \frac{i}{\hbar} [\hat{H}, \hat{f}]$ comes from differentiating $f(t) = \int dq \Psi^*(q,t) \hat{f}(t) \Psi(q,t) = \int dq \Psi^*(q,0) e^{i \hat{H}t/\hbar} \hat{f}(t) e^{-i \hat{H}t/\hbar} \Psi(q,0)$, commutators are unavoidable.
The Taylor expansion of $[\hat{H}, \hat{f}]$ in $\hbar$ about $\hbar = 0$ must be $0$ at order zero, and the PB at order $\hbar$ so that in the limit we get classical mechanics (which is defined as the $\hbar \to 0$ limit).
It's possible that the value of the PB can change if you have constraints, where it gets modified into the Dirac bracket, but that's separate from what your question is about
Randomly changing something in this picture makes no sense, but I bet you will get answers saying otherwise
 
@bolbteppa oh shit. i didnt consider this. so if I want the Born rule, I am enforcing the commutator as the bracket, right!!!?
 
I guess so yeah
 
@bolbteppa i was thinking I was getting the classical limit anyway with this alternative bracket.
@bolbteppa becuz the Heisenberg's eqns worked out fine by construction
 
@ACuriousMind If you want to win 300 internet points, see the challenge in the BFV comment I left on QMech's comment on my post
 
@naturallyInconsistent the reason we consider the commutator of fields is that observables like momentum, energy, angular momeentum are functions of the field variables
so the Feynman propagator is irrelvant for causality. position is not an observable
@bolbteppa hi again. i now think the commutator is really getting enforced by the Schrodinger eqn in ur derivation. but the Schrodinger eqn does not hold in the alternative quantisation i gave
the alternative quantisation assumes the Born rule.holds, and that the Heisenberg eqn holds @bolbteppa
 
7:37 AM
From $\psi \approx e^{iS/\hbar}$ (recall $S = \int (p dq - H dt)$) you have $\partial_t \psi = \frac{i}{\hbar} (\partial_t S) \psi = - \frac{i}{\hbar} H \psi$, thus the time derivative of the wave function has to reduce to this in the quasi-classical limit, fixing the Schrodinger equation, again unavoidable
 
@bolbteppa i am getting the classical limit here using the Heisenberg eqn. so the classical limit does hold. but the path integral of the action does not hold becuz it's equivalent to the Schrodinger eqn
 
Note sure what that means, but if you're doing something different you're basically not doing QM any more
 
yes, i know it's experimentally incorrect. only the Heisenberg eqn holds here
i am also looking for mathematical motivations to reject this, if they exist @bolbteppa
it does give u classical mechanics but the quantum mechanics here is incorrect
@bolbteppa i think there must be something special about the commutator as the Lie bracket?
 
Maybe Stone's theorem is the mathematical way to say what I said
 
stone's theorem does derive a wavefunction evolution eqn. so my Schrodinger picture also must follow some Schrodinger eqn according to Stone's theorem
and from there u can get the commutator
this is a great lead
 
7:48 AM
As far as I remember it says $\psi(q,t) = e^{iKt} \psi(q,0)$ where $K$ is unspecified, so you get commutators involving this $K$, how do we fix $K$, we do what I did above, if you don't do that you're not doing QM any more
 
yes, so even in this alternative quantisation, the operator evolution can be written in terms of a commutator
 
@SillyGoose The standard lore is that the mechanics of particles are just 1-dimensional field theory (the fields then are the trajectories $q(t)$ of the particles). But the grand picture you're looking for here does not exist: Even classical field theory is a mess if we want to state it fully mathematically because the "space of all fields" we want to talk about is infinite-dimensional, and no one knows the correct formalization of QFT in general at all.
 
the only problem is that now we have two operator evolution methods that r both compatible
this $K$ so far does not co incide with the Hamiltonian of this alternative quantisation @bolbteppa
 
so it sounds like the math is a mess... o.0
 
the Hamiltonian here is defined in terms of the other operator evolution eqn
 
7:51 AM
The only reason this happens is because you are assuming the superposition principles which implies linearity which implies Stone, if you are going to ignore the classical limit then there's no reason to even assume linearity because you've already given up trying to describe reality
 
@ACuriousMind I thought we would cheat by having Dirac delta distributions for the field strengths? I know that the formal equations would seriously throw a fit and break down, but I didn't know it cannot even be done this way?
 
@bolbteppa I'm fully convinced the pointers to Henneaux' treatment of the question are correct and solve the issue fully; I'm not interested to play in this strangely restricted world where we apparently must derive BRST without using its well-established theoretical background
@naturallyInconsistent Which part of what I said do you think this cheat invalidates?
 
You're convinced of it but you can't explain why (and any attempt will fall apart and end up pulling rabbits out of a hat at some intermediate step) and neither can the answer there :p
 
@ACuriousMind I dont actually know?
 
@bolbteppa I've already tried to explain it to you, but this dismissive attitude is exactly why I will not try any further
 
7:56 AM
oh no more drama anxiety
 
He should use the dismissive attitude he shows on such things, onto his beloved string theory
@Relativisticcucumber hugs
 
@bolbteppa how do you define reality?
sounds like a rabbit out of a hat to me 0.0
 
people r mass downvoting my question
it's a genuine question about foundations. idk y everyone is so weird
 
Speaking of rabbits in hats and reality
 
8:05 AM
i cant even upvote it myself...
stackexchange undoes it automatically
i think it will get to -5
i had never gotten that low
@bolbteppa QMechanic commented that $[X,P]=0$ is the other option. is there a proof of this
The requriment is that both X and P shud hav unbounded continuous spectrum
 
I think the comment meant, are you trying to assume that randomly, not that it is consistent to do this, again remember the PB of $x$ and $p$
 
assume what?
im not assuming [X,P]=0, but my requirements dont rule this out
@bolbteppa i think [X,P]=0 is a consistent theory after all
and this also circles back to what we werr just discussing. in this theory, the Hamiltonian of the Schrodinger eqn does not line up with the Hamiltonian of the alternative Lie bracket @bolbteppa
i shud add this to the post
@bolbteppa this theory is called KvN mechanics. it's just classical mechanics
i think we r getting somewhere. hopefully [X,P]=0 and [X,P]=i are the only two good possibilities
this shows that both the alternative Lie bracket and the commutator can co-exist in a theory. their evolutions are compatible
and in QM, these two brackets are the same thing
 
8:30 AM
Well it's not 'quantizing' when you do this, you're not assuming $\psi \approx e^{iS/\hbar}$, and it's very ugly, but it's probably what you're looking for
 
Commutator at zero is the Koopman-von Neumann theory
 
If you take this seriously you are probably going to have non-relativistic $SO(3)$ spinors even in the Kepler problem, the wiki mentions them arising in the relativistic case (presumably the writer thinks relativity weirdness makes it okay they arise)
 
in this theory (above) we have a conserved "charge" (second picture). in what sense is this a "charge"? and what is the symmetry associated with this quantity being conserved? i think this charge is general like how we have conserved currents, but i still dont see what the symmetry is
 
@Relativisticcucumber It's the $\mathrm{U}(1)$ symmetry $\phi\mapsto \mathrm{e}^{\mathrm{i}\alpha}\phi$ for any $\alpha\in\mathbb{R}$
 
@Relativisticcucumber The complex scalar field is "famously" a scalar field that can interact with the electromagnetic field because it can be imbued with some electromagnetic charge.
 
8:43 AM
you don't need to couple the field to EM to have this global symmetry
 
@bolbteppa i added this answer physics.stackexchange.com/a/792026/156987
 
@ACuriousMind Just to be exceedingly clear, this is the global phase, not a local phase
 
if you couple it, then this Noether charge will also become electric charge, but that's a rather silly statement, because without it, this charge is the net particle number, and of course if you assign an integer charge to particles whose net number is conserved, then that charge is also conserved
 
@naturallyInconsistent hmmm did u just downvote it for no reason?
i just posted it. there's no way anyone read it full
 
@RyderRude Please don't accuse specific other users of downvoting you, especially if they haven't visibly interacted with your post at all; votes are intentionally private
 
8:51 AM
ok. it's just that they've said before that they want to downvote my other post. i will not do it again.
 
If you think someone is targeting you unfairly with votes, please raise a moderator flag - bringing accusations like this up publicly serves no purpose since ordinary users cannot see anything about other users' voting patterns, while moderators have tools to investigate this kind of thing
 
ok. i m really sorry if i was wrong (and sorry in general too) @naturallyInconsistent
 
@ACuriousMind what is the physical interpretation of this symmetry and the conserved charge associated with it? and it doesnt relate to the EM field at all? but U(1) is the symmetry for electromagnetism, right?
 
@ACuriousMind what is the difference between a global and local symmetry?
 
In this case, I did downvote it, but I had good reasons. When people make both questions and answers, it is rare for them to be good, like Emilio Pisany's recent wonderful Q&A on the Nobel prize work. In your case, you posted quite a lot of the preliminary work here in this chat, so it was very easy to verify that the answer is problematic. In particular, you asked for a general question on whether a different commutator choice would be logically inconsistent,
 
9:00 AM
@Relativisticcucumber In the context of the theory of just a scalar field, this symmetry "rotates particles into antiparticle" (look at what it does to the c/a operators) and the conserved charge is, as I already said, net particle number, as any text that writes $Q = N_c - N_b$ should explain
 
but answered by only considering i and 0; i.e. your answer did not even answer your stated question.
 
it literally showed that the 0 case is consistent!!???
 
$N_c$ is the number of particles, $N_b$ the number of antiparticles, their difference is "net particle number", and it being conserved means you can only ever create or destroy a particle if you simultaneously create or destroy an antiparticle
 
and there's no way u had read it. the downvote was instant
 
@SillyGoose one is global, the other is local? :P
 
9:02 AM
@naturallyInconsistent the fact that the first reason u give is "it's rare for both of them to be good" speaks a lot
thanks for admitting it
 
hm okay i think i can accept that for now. hopefully i see more when moving to not scalar fields
 
@SillyGoose more seriously, see this answer of mine
 
@Relativisticcucumber Actually, I think, this time, you should not. Once you throw in spinors, it would get more troublesome. It would be illuminating to see how this charge-ness interacts with the EM field when you upgrade the global symmetry into a local one.
 
we don't really "upgrade symmetries" :(
I don't know who first thought it was a good idea to talk about gauge symmetries in that way but it's a terrible pedagogical choice that obscures the radically different character of a gauge symmetry and it doesn't even make sense because theoretically you can have arbitrarily many U(1) symmetries and the $\phi$ is charged differently under each one (or even uncharged) and it still also has the global symmetry from before
 
9:10 AM
I agree with you, ACM, it is just that the usual texts would phrase it that way, so I'm kinda forced to a life of sin
Hey, ACM, since we are at this, what happens if we wanted to avoid the gauge invariant Lagrangian theory and instead try to do physics with the constrained Hamiltonian? Would the resulting theory be extremely annoying to work with?
 
@naturallyInconsistent I mean - what exactly do you want to do with the theory?
the equations of motion of both formalisms are just Maxwell's equations
most of classical EM is about solving those, not doing stuff with the Lagrangian or Hamiltonian formalisms themselves
and when we quantize it, we're implicitly using the Hamiltonian formalism anyway
 
That much is certain, but I wanted to know if the derivations and the "how to avoid accidentally violating gauge symmetry" would be much more complicated
 
QFT can also be done after gauge fixing
 
Because maybe it would be a slightly simplier logical flow for students to follow, but end up with a tremendously more difficult to avoid accidentally doing something wrong and throwing the entire computation into a dungeon
 
@naturallyInconsistent you don't have to pay attention to "not violating gauge symmetry" in the Hamiltonian formulation, you just have to pay attention to enforcing the constraints
but the main constraint is just Gauß' law :P
there really isn't a lot to the Hamiltonian formulation here unless you quantize
 
9:19 AM
Oh, if this is just it, then why aren't we teaching it this way?
 
again - isn't almost everything we teach in EM concerned with solving Maxwell's equations rather than the Lagrangian or Hamiltonian formulation specifically?
I'm not sure at what point you imagine we should do something differently, here
sure, people could talk about the Hamiltonian equivalent of the Lagrangian in the very brief section where we talk about the Lagrangian of EM, but then you'd have to develop the entire theory of Hamiltonian constraints just to state what's going on...and end up with Maxwell's equations again
 
In the current canon of teaching QFT, quite a lot of emphasis is upon the importance of gauge symmetry and how they describe interactions. That is part of why we have all these upgrading of symmetries talk that annoy you. If we could just sidestep all of that, we could just so straight to Hamiltonians with constraints that then would make it less likely to bring up the upgrading of symmetries talk.
 
oh, but if you do Hamiltonian constraints you then have to teach BRST instead of hacking your way through the problem with Gupta-Bleuler to get to QED
 
See, I knew something is going to be worse, or else we would already have had many treatments doing that.
Will I turn into a bolbteppa, railing against BRST, if I went down that route?
 
I agree, in my ideal world that is what we would do, but this requires a lot more prerequisites - usually people coming to QFT fresh have never heard of constrained Hamiltonian mechanics at all
 
9:31 AM
Not at all; they would have heard vague notions that the SR or GR scalar-invariant Hamiltonian is identically zero being cryptically thrown around
 
and if you don't develop this theory carefully, the result will be an even greater mess than what you already get; it's fine to do only the Lagrangian + Gupta-Bleuler hack in a first approach to QED
but that doesn't require talking about "upgrading the symmetry" - in my view, we should start not from the observation that the complex fields have a global U(1), but from the observation that free EM has the local U(1), and that we're not allowed to break that (either by an anomaly argument or by a "gauge symmetry protects the mass" argument), so when we want to couple this to scalar fields, it is allowed to non-trivially act as a local U(1) on those, too
 
"Constrained mechanics" pre-QFT usually means "Particle on a ring"
By which I mean "a marble on a stick"
 
the argument for why we're "not allowed to break" the gauge symmetry will be a bit more involved than just upgrading the symmetry, but at least it has meaningful physical content: That masslessness of vector bosons and gauge symmetry are linked is an important principle
 
ACM, do you think it would be possible to make a pedagogical treatment that starts from EM field and quantising that, i.e. so we have local U(1) right from the start, and then insisting that the wavefunction of whatever we want to make electrons obey, also respect that local U(1) symmetry?
Because I kinda feel that, this way, we would lose the standard QM treatment of emphasising the unmeasurability of the global phase, or something like that.
 
@naturallyInconsistent that people confuse the U(1) of electromagnetism for the global phase irrelevance is another one of my pet peeves!
 
9:38 AM
Then could you put in some comments as to a better view? As you might remember, I'm trying to write a textbook, so if you make this clear, I might be able to find some way to incorporate it.
 
the two are entirely different: If you have three states of charge $0$, $e$ and $2e$, then the global phase irrelevance for $\lvert 0\rangle + \lvert e\rangle + \lvert 2e\rangle$ is $\mapsto \mathrm{e}^{\mathrm{i}\alpha}(\lvert 0\rangle + \lvert e\rangle + \lvert 2e\rangle)$ while the EM U(1) is $\mapsto \lvert 0\rangle + \mathrm{e}^{\mathrm{i}e\alpha} \lvert e\rangle + \mathrm{e}^{\mathrm{i}2e\alpha}\lvert 2e\rangle$
the reason this kind of confusion doesn't usually harm anything is that charge is often superselected for (we're in a space of constant, definite charge) so you can't see the interference terms the second transformation produces anywhere
 
Isn't it mandatorily superselected
 
Hmm, superselection is another annoying topic to think about
 
It's the fundamental rule of QM nobody talks about
 
@naturallyInconsistent I think the best way to get to QED is just to write down the classical EM Lagrangian with source current and say "let's assume this current comes from an underlying theory where $j^\mu$ is made from other dynamical fields", and since we know that there should be something that produces electrons here, we just make the simplest possible fermion-fermion-vector boson interaction that preserves gauge invariance
@Slereah I really didn't want the discussion to get sidetracked by pedantic arguments about what assumptions are necessary to prove this superselection :P
 
9:48 AM
What is the relation between the U(1) global symmetry and the U(1) of EM anyway
While they're somewhat independent it's still the U(1) charge that is the source for EM
Or is that not true I'm not sure
 
@ACuriousMind these two are extremely different symmetries in QFT, but in QM, the gauge field is indeed coupled to the phase of the wavfunction. so that the Schrodinger eqn becomes local phase invariant.
becuz now the schrodinger eqn gets the covariant derivative term
the probabilities also become local phase invariant in this theory
becuz the observables depend on gauge
so that the local phase transform transforms the observables too
like the observable momentum is P-ieA, whose expectation value is local phase invariant
 
@ACuriousMind What I am worrying about is the pedagogical learning curve's steepness. I'm trying to smoothen and flatting and make things simple and understandable. But I do think that it might be possible to do something like what you are suggesting.
 
I have small question regarding the Feynman propagator. Since
$\square_x\Delta(x-y)=-i\delta^4(x-y)$
and the propagator is symmetric under the interchange of x and y, shouldn't the RHS be antisymmetric under the exchange? But the delta function is clearly not.
 
@AkshatSharma The propagator is symmetric, the D'Alembertian is a 2nd derivative and thus also symmetric, why do you think there would be skew-symmetry?
 
The delta function is an even function
It is invariant under interchange of x and y
Since $\delta(x) = \delta(-x)$
 
9:58 AM
@Slereah The questioner explicitly noted that this is true at the end.
 
@naturallyInconsistent Ah right it's the second derivative. I was using the chain rule after the exchange so I was expecting antisymmetry.
 
yay problem solved
 
Thanks!
 
10:26 AM
Actually is it even true tho
What if you have multiple U(1) gauge fields
 
10:41 AM
@Slereah then you have multiple "electric charges"
nothing special
 
My point being that if a given object has multiple couplings to different U(1) fields, probably means that the phase U(1) symmetry charge doesn't relate directly to the U(1) gauge charges
I think anyway?
 
that's what I keep saying, isn't it? ;)
 
Is the EM charge just "There exist a gauge transformation that is invariant on the whole space that triggers the first Noether theorem" or something
 
well, that goes back to what I said above: The "global" part of all these different U(1) symmetries acts the same on all charged objects, and the Noether charge of that is always just the net particle number operator with a pre-factor related to the charge
you don't really get "different" Noether charges for different U(1) symmetries in this sense
 
i got an upvote but only +2 got added to my score
 
10:48 AM
Up to some factor anyway
 
@RyderRude you didn't get an upvote, someone removed their downvote (click on the -3 to see that the score is 0/-3, not +1/-4)
 
yes. i just figured it out
@ACuriousMind can we pls be on good terms again
 
Any one of the whole list of us would be willing and able to be on good terms with you; the ball is in your court.
 
it wasnt for u
 
@RyderRude Maybe re-phrase it as asking 'is it possible to perform canonical quantization without assuming $[x,p] = i$ and what do you get?' Am I wrong or is your Poisson bracket going to give $(x,p) = i$ even though your post is saying we should not assume $(x,p) = i$? Not clear how KvN is an answer to the question you asked. Looks like you're asking if we can do QM with the Poisson bracket, I would maybe take it out altogether
 
10:57 AM
@naturallyInconsistent sorry i will try to be on good terms with u too
@bolbteppa i think the question is fundamentally about two different Lie brackets defined on operators on Hilbert space. when these two co incide, u get QM. when these two dont co incide, i want to explore some generic situations
one of the cases when they dont co incide is KvN
 
But you've written the normal Poisson bracket which satisfies $(q,p) = i$ while saying we can't assume this, it's just very confusing as written, think of the reader
 
but i want to know how to reject the other possibilities
@bolbteppa yes. this alternative lie bracket will always satisfy (q,p)=i. it's the commutator Lie bracket that can fail to satisfy this, e.g. in KvN
 
But you said we can't assume this, you're just confusing everyone
 
the bracket i've written is the same as the QM bracket only when [q,p]=i holds
@bolbteppa what im not assuming is the value of the commutator. the Lie bracket ive defined is not the commutator of the operators
i have given an independent definition
 
You defined a commutator for $f$ and $g$, if I set $f = X, g = P$, I get $(X,P) = i$ which you said we can't assume, it doesn't make any sense as written, or maybe it does but you could make it clearer
 
11:04 AM
but note that u did not have to assume X=x, and P=-id/dx to get this result from my definition
 
But you began by stating we can't assume $(x,p) = i$ framing the whole question as being about this then immediately ignore that and define things where $(x,p) = i$, how do you not see that
 
i said we cant assume [X,P]=i. (X,P)=i always holds in this quantisation
the title is about [X,P]=i
 
I'll leave it at that, I think what I said is clear, you need to make the question clearer if you want people to read it properly
 
i will re-phrase it. thanks
 
 
1 hour later…
12:16 PM
i will try to ask a new question. i want to ask something like "What is the mtmathematical motivation behind making the two Lie brackets on the Hilbert space equal"?
this equality leads to QM. a non-equal case leads to classical mech
and i want to know about the other cases, what they mean
 
Not sure what that means
 
let's say i have a hilbert space, and any two operators on it, X and P. now i can define two lie brackets on functions [f(X,P), g(X,P)] and (f(X,P),g(X,P)). these two have independent definitions
in QM these two Lie brackets are equal
the first bracket is defined as $[f,g]=fg-gf
the second bracket using partial differentiation of the functions, analogous to a Poisson bracket
does this make any sense @bolbteppa
in QM, these two brackets happen to be equal
 
The Hilbert space isn't functions on the phase space
 
yes, im defining the partial differentiation to act as if they were a function
in the end, it's just a bracket definition
this bracket is actually useful in KvN
 
Where is the Poisson bracket coming from? You're just defining it out of thin air, differentiating w.r.t. operators, forcing the $i$ or $i \hbar$ in front, and saying the result is the same for $x,p$. Where did it come from
KvN seems to be using commutators $[A,B] = AB - BA$ and invoking Stone
 
12:23 PM
the other bracket, $(f,g)=i(\frac{\partial f}{\partial X}\frac{\partial g}{\partial P}-\frac{\partial f}{\partial P}\frac{\partial p}{\partial X}$ @Slereah
this is defined on functions of operators
 
But hilbert space vectors aren't functions of x and p
It's already overreaching to say they're functions at all
 
f and g are operators, not vectors
 
KvN isn't differentiating w.r.t. operators either
 
the other bracket is also defined on operators
@bolbteppa traditional KvN does not mention it, but this bracket generates the correct dynamics of KvN
the dynamics being $\frac{dO}{dt}=-i (O,h)$. $h$ is not the KvN Hamiltonian here. () is this new bracket
the dynmaics are also generated by the commutator in KvN
these two dynamics are compatible
the commutator dynamics uses the KvN Hamiltonian
i have explained this in my answer. these two bracket dynamics are equivalent
pls dont leave me hanging. i can explain why this bracket works
@Slereah the differentiation is of f(X,P) wrt X and P, pretending that this is an ordinary differentiation. it is not the differentiation of a wavefunction.
i agree that the wavefunction is in general not a function of x and p
 
Looks like nonsense man, maybe it makes sense but I doubt it
 
12:36 PM
so far it is just a definition of a bracket, right? ur only question might be, why am i defining this bracket
i can explain that. this bracket gives u the correct dynamics both in QM and in CM
in QM, this is just equal to the commutator, so correct dynamics
in CM, this bracket just gives u the operator version of Hamilton's eqns, as it is just the Poisson bracket in operator form
so again correct dynamics
now in CM, there is another way to get correct dynamics using the commutator
these two evolutions r compatible
the question that's intriguing is, in general, the commutator and this bracket are not the same operation. in classical mechanics, they arent $[f,g]\neq (f,g)$
in QM, they are the same thing
this is only because [X,P]=i
 
12:51 PM
No reason this bracket will ever arise from anything or ever need to be used even if it makes sense
 
Found the Aristotelian physics bit :
> A given weight moves a given distance in a given time; a weight which is as great and more moves the same distance in a less time, the times being in inverse proportion to the weights. For instance, if one weight is twice another, it will take half as long over a given movement. Further, a finite weight traverses any finite distance in a finite time.
You can tell he is using weird ancient ideas because he is arguing against infinite weight and infinite lightness
Which I assume is against the existence of zero mass
 
what are you reading now
 
"On the Heavens" by Harris Turtle
It's not great by modern standards so far
Ptolemy's book was a lot more impressive imo
> I mean that if it is impossible for a thing to have come to be white, or a cubit long, or in Egypt, it is also impossible for it to be in process of coming to be any of these.
The three basic physical process
 
1:09 PM
@bolbteppa my only question is, does $[X,P]=i$ purely come from experiments? becuz we can get the classical limit and the dynamics right without this
X and P cud be any operators with unbounded continuous spectrum on any Hilbert space
KvN is evidence that u can get the classical limit and the dynmaics right without [X,P]=i, but there also remain other choices of theories that r neither quantum nor classical
i mean theories with $[X,P]\neq i,0$
 
> Then, as D is to B, so is E to some finite quantum
Aristotle already doing quantum mechanics
 
i think this problem is useless. the correct quantum theory is $[X,P]=i$
no need to think about useless theories
thanks everyone
the only choice for these useless theories is $[X,P]=\hbar f(X,P)$. this is extremely unnatural if $f$ is non trivial
if $f$ is constant, then it has to be imaginary for Hermitian X and P
 
1:28 PM
From $\{x,p\}_{P.B} = 1$ and $i \{A,B\}= [A,B]$ (depends on classical limit) we get $[x,p] = i$. Alternatively we can derive $p = - i \partial_x$ (from the classical limit) and show $[x,p] = i$. This shows that any bracket involving associative operators with the basic properties of the PB will be a commutator.
I have no idea how your partial derivative of operators bracket relates to this, but it likely does not make sense as it seems to contradict the proof in that last pdf but maybe not I don't know
 
yes. they have proved this. thanks
 
Without the classical limit you're basically stuck with $[A,B]$ as some abstract commutator without more information, KvN seems to be providing information in the case where $[x,p] = 0$ being forced to introduce random non-commutative operators (that implicitly satisfy $[x,p] = i$) to make things work, just ugly stuff
 
im trying to figure out how the other bracket fits into this proof
@bolbteppa KvN produces the correct dynamics with my bracket too tho. non commuative operators r not needed
my bracket only uses commuting operators in case of KvN for the dynamics
@bolbteppa but this proof says this bracket shudnt make sense
i dont get it. maybe this bracket doesnt satisfy the same properties as Poisson? the proof assumes this
 
That is basically how Dirac set up most of QM via this proof more or less
 
oh
i read Dirac also tried $[X,P]=i\frac{1}{\hbar}$
 
Aristotle also proves that other worlds do not exist which I did not expect
 
@bolbteppa i just figured out how the other bracket fits into this proof. this proof is looking for a general bracket on operators on a Hilbert space, while my bracket is undefined on general operators
 
Basically arguing that the elements go to the center of the universe and therefore if you had another world that shit would go down or something
 
@bolbteppa the other bracket is only defined on functions $f(X,P)$
this bracket is sufficient to set up a theory becuz we dont care about general operators, but only observabels
 
That sounds very wrong, I think this is not worth pushing any further to be honest
 
1:52 PM
yes
 
> This conclusion that local movement is not continued to infinity is corroborated by the fact that earth moves more quickly the nearer it is to the centre, and fire the nearer it is to the upper place. But if movement were infinite speed would be infinite also; and if speed then weight and lightness.
It is all so bad
 
2:09 PM
so in summary : the dynamics must be given by some bracket on operators. the commutator is the only such bracket that is defined on general operators, according to the proof. from there, u must have $[X,P]=i$
Dirac also considered $[X,P]=\frac{i}{h}$ for some reason. but he proves this was inconsistent
sorry it's not Dirac. it's Groenewold's theorem
this other possibility is dimensionally incorrect. idk y anyone considered it lol
sorry Groenewold's theorem is about $[f,g]=ih(f,g)_{PB}$. the other choice makes no sense
@Slereah Aristotle is a huge armchair physicist
 
2:57 PM
In this answer it is said that we need locality and Lorentz invariance to define the stress energy tensor. Why is that? I found an intriguing example in ACM's answer to that question also mentioned in this other post. But why is locality and Lorentz invariance a necessity?
@naturallyInconsistent lol. stupid "typo" :p
 
3:22 PM
@Sanjana If the theory is not local, then the result of $\delta Z/\delta g$ will not be a local function. If the theory is not Lorentz invariant, then $Z$ transforms non-trivially under Lorentz transformations and so the result of $\delta Z/\delta g$ will not transform as a 2-tensor under Lorentz transformations.
but you want your stress-energy tensor to be a local function that transforms as a 2-tensor
 
@ACuriousMind Why Lorentz covariance and not general coordinate invariance which seems stronger? I mean why do I want my stress tensor to transform as a Lorentz 2-tensor in contrast to a 2-tensor under GCT?
 
there's no meaning to "coordinate invariance" :P
 
note that tensors r defined without any reference to the metric
 
more seriously, you might be aware of the problem with different definitions of stress-energy tensors, where some of them are only pseudo-tensors
 
@ACuriousMind I mean why want $\tilde{T}_{\mu \nu}=\Lambda^{\alpha}_\mu \Lambda^{\beta}_\nu T_{\alpha \beta}$, instead of the corresponding relation with $\Lambda$s replaced by some jacobians corresponding to a generic non-linear coordinate transformation?
@ACuriousMind And also, why should the stress energy tensor should always be "local"? I mean mathematically it's fine ig, but physically why can't our stress energy tensor be bilocal or something?
 
3:30 PM
@Sanjana Question: 1. Is free EM invariant under such transformations? 2. Does free EM have a stress-energy tensor?
@Sanjana that's just not what we mean by "stress-energy tensor"
 
@ACuriousMind okay, so it's too stringent to do meaningful physics...
@RyderRude noted...but what's the point?
@ACuriousMind Why? I mean...there must be some physical insight as to why we shouldn't consider some non-local thing obtained by doing $\delta Z/ \delta g$ a valid stress tensor...I mean blindly I can just take the derivative and say "I got a non-local stress tensor...this is something new, we should think about it", but instead I say "it's not a stress tensor...it's garbage"
why?
 
@Sanjana yes, you could say that
unless you know a practical application of non-local theories where you can demonstrate that your definition is still a meaningful stress-energy but is ignored by AFT's insistence on locality, I'm not sure why you'd get hung up on the word choice
 
Oh okay
@ACuriousMind I know only one example---the Landau-Lifschitz pseudotensor
 
@Sanjana is there no stress energy tensor for the Schrodinger lagrangian? it's local and non relativistic
i think it shud hav it becuz it models non relativistic QFT
it ofc has a Hamiltonian density
so i dont think relativity is a requirement for this
 
Aristotle's argument as to why the celestial's sphere motion is naturally circular :
> Such a constrained movement would necessarily involve effort the more so, the more eternal it were-and would be inconsistent with perfection.
 
3:43 PM
I love a good old "the world would suck if that was the case, so it isn't" argument
 
I mean if I was the creator of the universe I wouldn't want to have to spin the celestial sphere by hand all the time
 
why not? what else are you doing with your time?
 
If going by Greek myth quite a lot
You know there's a weird thing in the bible where one of the book ends abruptly and then the following book starts exactly at that point again and then I learned that this is because of the length of torah scrolls
I wonder if something like that is going on with greek books too
On the Heavens is just split into 4 books without much thematic changes in between each book
 
@RyderRude Yes, I don't see Lorentz covariance as a necessity. The locality aspect is what I get to see in places most of the times...
 
@Slereah the Greek probably wrote mostly on papyrus, too
so it's not unlikely their scrolls had similar length restrictions
 
3:50 PM
@Sanjana yes. the locality part was intuitively expected so i just accept it. i wud b worried if non local theories allowed for an energy density
 
@RyderRude Why would you expect that non-local theories won't have energy density?
 
i hate non intuitive results @Sanjana
 
@Sanjana Non-relativistic theories may have a stress tensor, but not stress-energy tensors, that's just how we use the word. Don't get so hung up on minor issues of phrasing.
 
@Sanjana the locality ---> Density thing is expected. i dont see an obvious way to see this the other way around. so i wudnt expect it intuitively, but i probably wud expect it anyway because my intuition is wrong usually
 
I am not bothered much about the Lorentz covariant part.
I am actually not fully convinced by why we shouldn't expect a stress energy tensor for non-local field theories?
 
3:57 PM
Context: Goldstone theorem. What is the meaning of this sentence?
 
@Sanjana Can you name a non-local field theory? Can you point out something in that theory that you'd like to call stress-energy?
 
basically: what the hell does it mean that a state has the same quantum numbers as an operator
 
I find it deeply pointless to try to generalize concepts beyond how they're usually used if one does not have at least one example for where this would be useful
@Mr.Feynman If $[Q,A] = qA$, then $A\lvert 0\rangle$ is an eigenvector of $Q$ with eigenvalue $q$.
and purely formally $A$ is an eigenvector of $[Q,-]$
 
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